Answer:
The charges under study are of the same sign
The calculation of the electric field for each charge separately, there is no relationship between the charges
Explanation:
Let's start by writing the equation for the electric field
E = k q / r²
where q is the charge under analysis and r the distance from this charge to a positive test charge.
When analyzing the statement the student has some problems.
* The charges under study are of the same sign, it does not matter if positive or negative.
* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.
* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero
Answer:
2.5 * 10^-3
Explanation:
<u>solution:</u>
The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:
<em>δu/δx+δv/δy=0</em>
so that:
<em>δv/δy= -δu/δx</em>
Now, since u = Uy/δ, where δ = cx^1/2, we have that:
<em>u=U*y/cx^1/2</em>
and we obtain:
<em>δv/δy=U*y/2cx^3/2</em>
The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):
v=∫δv/δy(dy)=U*y/4cx^1/2
=y/x*(U*y/4cx^1/2)
=u*y/4x
which is exactly what we needed to demonstrate.
Also, using u = U*y/δ in the last equation we can obtain:
v/U=u*y/4*U*x
=y^2/4*δ*x
which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:
(v/U)_max=δ^2/4δx
=δ/4x
=2.5 * 10^-3
The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing 
for n and 202Hz for 
The frequency of the tightened is 205Hz
Answer:
Both are attractive as well as repulsive.
Explanation:
(Like poles repel, like charges<em> repel</em>; unlike poles attract, unlike charges <em>attract</em>).
Answer:
A trough
Explanation:
A trough is an elongated area of relatively low pressure extending from the center of a region of low pressure.
I HoPe ThIs Helps!!!
